Question: The equation of a circle $C$ is $x^2+y^2-16x-4y+59 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-16x) + (y^2-4y) = -59$ $(x^2-16x+64) + (y^2-4y+4) = -59 + 64 + 4$ $(x-8)^{2} + (y-2)^{2} = 9 = 3^2$ Thus, $(h, k) = (8, 2)$ and $r = 3$.